An $AP$ consists of $37$ terms. The sum of the three middle most terms is $225$ and the sum of the last three is $429$. Find the $AP$.

(A) Given,total number of terms $n = 37$.
The middle term is the $\left(\frac{37+1}{2}\right)$-th term,which is the $19$-th term.
So,the three middlemost terms are the $18$-th,$19$-th,and $20$-th terms.
According to the given condition,the sum of the three middlemost terms is $225$:
$a_{18} + a_{19} + a_{20} = 225$
$(a + 17d) + (a + 18d) + (a + 19d) = 225$
$3a + 54d = 225$
$a + 18d = 75$ .....$(i)$
The sum of the last three terms is $429$:
$a_{35} + a_{36} + a_{37} = 429$
$(a + 34d) + (a + 35d) + (a + 36d) = 429$
$3a + 105d = 429$
$a + 35d = 143$ .....$(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(a + 35d) - (a + 18d) = 143 - 75$
$17d = 68$
$d = 4$
Substituting $d = 4$ into equation $(i)$:
$a + 18(4) = 75$
$a + 72 = 75$
$a = 3$
The required $AP$ is $a, a+d, a+2d, a+3d, \dots$
Substituting the values,we get: $3, 3+4, 3+2(4), 3+3(4), \dots$
Thus,the $AP$ is $3, 7, 11, 15, \dots$